Calculus of a Single Variable, Chapter 9, 9.4, Section 9.4, Problem 17

Limit comparison test is applicable when suma_n and sumb_n are series with positive terms. If lim_(n->oo)a_n/b_n =L where L is a finite number and L>0 , then either both series converge or both diverge.
Given series is sum_(n=1)^oo(2n^2-1)/(3n^5+2n+1)
Let the comparison series be sum_(n=1)^oon^2/n^5=sum_(n=1)^oo1/n^3
a_n/b_n=((2n^2-1)/(3n^5+2n+1))/(1/n^3)
a_n/b_n=(n^3(2n^2-1))/(3n^5+2n+1)
a_n/b_n=(2n^5-n^3)/(3n^5+2n+1)
lim_(n->oo)a_n/b_n=lim_(n->oo)(2n^5-n^3)/(3n^5+2n+1)
=lim_(n->oo)(n^5(2-n^3/n^5))/(n^5(3+(2n)/n^5+1/n^5))
=lim_(n->oo)(2-1/n^2)/(3+2/n^4+1/n^5)
=2/3>0
The comparison series sum_(n=1)^oo1/n^3 is a p-series with p=3
As per p-series test sum_(n=1)^oo1/n^p is convergent if p>1 and divergent if 0Since the comparison series sum_(n=1)^oo1/n^3 converges, so the series sum_(n=1)^oo(2n^2-1)/(3n^5+2n+1) as well ,converges by the limit comparison test.